Show that $B(1-t) - B(1)$ is a Brownian motion I need a check on the following exercise. In particular, I'd like to be sure that points ii), iii) are okay.

Let   $(B_t)_{t \in [0,1]}$ be a standard Brownian motion. Prove that the process $W_t=B(1-t) - B(1)$ defined for $t \in [0,1]$ is a standard Brownian motion


i) First I can see that $W(0)=B(1) - B(1)=0$ a.s.
ii) I need to check the the increments are independent. To this aim, I fix a partition of $[0,1]$, $0<t_1 < t_2 < \ldots < t_n$ and consider the increments $(W_{t_1}, W_{t_2} - W_{t_1}, \ldots, W_{t_n} - W_{t_{n-1}})$
I have that $$W_{t_{i}} - W_{t_{i-1}} = B(1 - t_i) - B(1 - t_{i-1})$$
and $$W(t_{i+1}) - W(t_i) = B(1 - t_{i+1}) - B(1-t_i)$$
I need to check they're independent: since both of them are normally distributed, to check the independence is enough to check that the covariance is $0$. Moreover, since $B$ is a Brownian motion, I can use the fact that 
$Cov(B_t, B_s)= s \wedge t$.
Hence $$Cov(B(1-t_i) - B(1-t_{i+1} ), B(1-t_{i-1})  - B(1-t_i)) = \\
(1-t_i) \wedge (1-t_{i-1}) - (1-t_{i+1}) \wedge (1-t_{i-1}) - (1-t_i) \wedge (1-t_i) + (1-t_{i+1}) \wedge (1- t_i) = 0$$
Hence the increments are independent.
iii) I need to show that for evert $0\leq s <t$: $W(t) - W(s) - N(0,t-s)$
I have that $W(t) - W(s) = B(1-t) - B(1-s)$. Now I have that $1-t< 1 -s$, therefore I note that $$ B(1-t) - B(1-s) = -(B(1-s) - B(1-t)) $$
Now I see that the right hand side has normal law since it's an increment of the Brownian motion. Moreover, I have that it's mean is clearly $0$, while the variance is $t-s$
Therefore $W(t) - W(s) - N(0, t-s)$.
(iv) Path continuity
I have that $t \mapsto B_t(\omega)$ is continuous for $P$-almost every $\omega$. Hence, the function $$t \mapsto B_{1-t}(\omega) - B(1)$$ is continuous, since is obtained from $B_t$ by a traslation and adding a constant $B(1)$.
Edit
To show it's a  Gaussian process, I can note the following:
first I note that $(B(1-t_n), B(1-t_{n-1}), \ldots, B(1-t_1), B(1))$ is a Gaussian vector (since $B$ is a Brownian motion). Therefore, I can see
$\begin{pmatrix} W(t_1) \\ \vdots \\ W(t_n) \end{pmatrix}=T((B(1-t_n), B(1-t_{n-1}), \ldots, B(1-t_1), B(1)))$
where $T$ is the linear map $T(x_1,x_2, \ldots,x_n)=(x_{n-1} - x_n,x_{n-2} - x_n, \ldots, x_1 - x_n )$.
Hence, since it's the image via a linear map of a Gaussian vector, $\begin{pmatrix} W(t_1) \\ \vdots \\ W(t_n) \end{pmatrix}$ is a Gaussian vector.
 A: Your proof looks good for the most part. However, it should be noted that two random variables being normal and having zero covariance does not imply independence. Wikipedia even has a nicely titled page for this.
Your calculation can be justified, however. The key point is that the standard Brownian motion taken at any two times $s,t$ produces two jointly Gaussian variables. It suffices to show that two jointly Gaussian variables are uncorrelated to show independence.
EDIT: Justification of independent increments.
Consider any two disjoint increments $[s_1, t_1]$ and $[s_2, t_2]$ of the process $(X)_t = B_{1-t} - B_1$. We want to show that $Y_1 = X_{t_1} - X_{s_1} = B_{1-t_1} - B_{1-s_1} $ is independent of $Y_2 = X_{t_2} - X_{s_1} = B_{1-t_2} - B_{1-s_2} $. It's actually easier to forego the covariance calculation. 
$Y_1 = B_{1-t_1} - B_{1-s_1}$ is the random variable tracking the Brownian motion in the time period $[1 - t_1, 1 - s_1]$ (we've flipped the endpoints as the negative of a standard Brownian motion is also a standard Brownian motion). Similarly, $Y_2 = B_{1-t_2} - B_{1-s_2}$ is the random variable tracking the Brownian motion in the time period $[1 - t_2, 1 - s_2]$. Since $[s_1, t_1]$ and $[s_2, t_2]$ are disjoint by assumption, the intervals $[1 - t_1, 1 - s_1]$ and $[1 - t_2, 1 - s_2]$ are also disjoint, and hence by the independent increments property of the standard Brownian motion, $Y_1 = B_{1-t_1} - B_{1-s_1}$ and $Y_2 = B_{1-t_2} - B_{1-s_2}$ must be independent too.
